Help simplifying expression I have the following set of equations:
1)
$\text{$b_1$} \cos (\text{$\beta_1$} u)-\text{$b_1$} \cosh (\text{$\beta_1$} u)-\text{$b2$} \cos (\text{$\beta_1$} \theta  y)-\text{$d_2$} \cosh (\text{$\beta_1$} \theta  y)\sin (\text{$\beta_1$} u)-\sinh (\text{$\beta_1$} u)=0$
2)
$-\text{$b_1$} \sin (\text{$\beta_1$} u)-\text{$b_1$} \sinh (\text{$\beta_1$} u)+\text{$b_2$} \theta  \sin (\text{$\beta $1} \theta  y)-\text{$d_2$} \theta  \sinh (\text{$\beta_1$} \theta  y)+\cos (\text{$\beta_1$} u)-\cosh (\text{$\beta_1$} u)=0$
3)
$-\text{$b_1$} \cos (\text{$\beta_1$} u)-\text{$b_1$} \cosh (\text{$\beta_1$} u)+\alpha ^4 \text{$b_2$} \theta ^2 \cos (\text{$\beta_1$} \theta  y)-\alpha ^4 \text{$d_2$} \theta ^2 \cosh (\text{$\beta_1$} \theta  y)-\sin (\text{$\beta_1$} u)-\sinh (\text{$\beta_1$} u)=0$
... and I would like to find the constants $b_1$, $b_2$ and $d_2$.
I already know the answers:
$b_1 =\frac{\sin \left(\beta _1 u\right)-\sinh \left(\beta _1 u\right)}{\cosh \left(\beta _1 u\right)-\cos \left(\beta _1
   u\right)}$
$b_2 =\frac{2 \cos \left(\beta _1 u\right) \left(\cos \left(\beta _1 u\right) \cosh \left(\beta _1
   u\right)-1\right)}{\theta  \left(\cosh \left(\beta _1 u\right)-\cos \left(\beta _1 u\right)\right) \left(\cos
   \left(\beta _1 \theta  y\right) \sinh \left(\beta _1 \theta  y\right)+\sin \left(\beta _1 \theta  y\right) \cosh
   \left(\beta _1 \theta  y\right)\right)}$
$d_2 = -b_2 \frac{\cos \left(\beta _1 \theta  y\right)}{\cosh \left(\beta _1 \theta  y\right)}$
I want to find those expressions myself, but so far I only have that (using Mathematica): 
Click on it, to see it full-scale...

Any help would be very much appreciated !
 A: Writing the system as
$$Ab_1+Bb_2+Cd_2+D=0\tag1$$
$$Eb_1+Fb_2+Gd_2+A=0\tag2$$
$$Hb_1+Ib_2+Jd_2+E=0\tag3$$
should make it easy to solve where
$\qquad\begin{cases}A=\cos (\text{$\beta_1$} u)-\cosh (\text{$\beta_1$} u)&
\\\\B=-\cos (\text{$\beta_1$} \theta  y)&
\\\\C=-\cosh (\text{$\beta_1$} \theta  y)\sin (\text{$\beta_1$} u)&
\\\\D=-\sinh (\text{$\beta_1$} u)&
\\\\E=-\sin (\text{$\beta_1$} u)-\sinh (\text{$\beta_1$} u)\end{cases}$ $\qquad\begin{cases}F=\theta  \sin (\text{$\beta $1} \theta  y)&
\\\\G=-\theta  \sinh (\text{$\beta_1$} \theta  y)&
\\\\H=-\cos (\text{$\beta_1$} u)-\cosh (\text{$\beta_1$} u)&
\\\\I=\alpha ^4 \theta ^2 \cos (\text{$\beta_1$} \theta  y)&
\\\\J=-\alpha ^4 \theta ^2 \cosh (\text{$\beta_1$} \theta  y)\end{cases}$

$E\times (1)-A\times (2)$ gives
$$Kb_2+Ld_2+M=0\tag4$$
$H\times (1)-A\times (3)$ gives
$$Nb_2+Pd_2+Q=0\tag5$$
where
$$K=BE-AF,\quad L=CE-AG,\quad M=DE-A^2$$
$$N=BH-AI,\quad P=CH-AJ,\quad Q=DH-AE$$
$P\times (4)-L\times (5)$ gives
$$(PK-NL)b_2+PM-QL=0,$$
i.e.
$$b_2=\frac{QL-PM}{PK-NL}$$
From $(4)$, 
$$d_2=\frac{MN-KQ}{PK-NL}$$
Finally, from $(1)$, we get
$$b_1=\frac{-B(QL-PM)-C(MN-KQ)-D(PK-NL)}{A(PK-NL)}$$
